Ask/Help sape terer add math masuk sini

[adijaya]

soalan 2:http://latex.codecogs.com/gif.latex?x^2%20e^4^x

jawapan untuk soalan 2 plak camni:



jawapan dia panjang skit untuk soalan ni.
kena ganti banyak step sebab kamiran dia terhadap e kuasa empat x dx,
dimana e kuasa empat x adalah exponent.

 

[kaYaIlmu]

aku dapat A1 addmath masa spm dlu..tapi kalau tanya sekarang ,langsung aku xingat ...yang aku ingat adalah MONEY ...wakakak..

 

[adijaya]

bro pensyarah kat mana?

saya bukan pensyarah pun. huhu.
sekadar membantu setakat yang termampu.

 

[RuthlessFairy]

) )

add math....

aku paling benci buat latihan.
bila cikgu bagi kerja sekolah, aku tak penah buat by my own,
aku selalu tender copy sorang budak cina nama dia ki huat yee.


tapi spm add math aku dapat kredit.
cukuplah untuk melayakkan aku amek arhitecture.

tapi jgn tanya aku. aku lupa.
sbb aku study utk exam je.

 

[aimanxxx]

answer is 1.

 

[ezxury]

thx2 selesai

 

[abahmu]

otak dh berkarat..x ingat ape2 wlupun spm dulu dpt A1 )

 

[Megat Sri Rama]

Nah refer sini..))

http://www.youtube.com/watch?v=8uFQeJiXC5M

macam mana boleh sama dengan soalan tu?

betul la tu jawapan dia.

 

[Megat Sri Rama]

zaman sekarang senang kalau nak belajar...

semua benda ada kat internet.

seingat aku sampai tahun 2000 dulu pun masih kurang tutorial kalkulus kat internet.

 

[epol rider]

The idea is to integrate by parts twice. Start with the original integral,
∫sin(x)e^(2x) dx =

Let u = e^(2x) and dv = sin(x). Then du = 2e^(2x) and v = -cos(x). Apply integration by parts to get:

∫u dv = uv - ∫v du
∫e^(2x)sin(x) dx = -e^(2x)cos(x) - ∫-cos(x)2e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx

Call this last equation *1*. Now this seems to have made the integral we couldn't compute, ∫e^(2x)sin(x) dx, into yet another integral we can't compute, namely ∫cos(x)e^(2x) dx. So, apply integration by parts again - this time on just that rightmost integral.

For ∫cos(x)e^(2x) dx, let u = e^(2x) and dv = cos(x). Then du = 2e^(2x) and v = sin(x). So, from integration by parts you get:

∫u dv = uv - ∫v du
∫e^(2x)cos(x) dx = sin(x)e^(2x) - ∫sin(x)2e^(2x) dx
∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2∫sin(x)e^(2x) dx

Now plug this substitution of the integral ∫e^(2x)cos(x) dx into equation *1* and see if it becomes cleaner.
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2[sin(x)e^(2x) - 2∫sin(x)e^(2x) dx]
∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x) - 4∫sin(x)e^(2x) dx
∫e^(2x)sin(x) dx + 4∫sin(x)e^(2x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)
5∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))
∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))/5

Hope this helps you!

satu benda pun aku tak paham

aku dulu ambil sastera je.Jadi Mat Jiwang...
kalau TT nak minta buat puisi cinta ka,bolehlah aku tolong )